In right angle triangle ABC right angled at P and Q are points of sides CA and CB respectively which divide there in the ratio 2:1 prove that 9AQsq =9ACsq+4BCsq
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Answer :-
In ACQ
(1)AQ
2
=AC
2
+QC
2
QB
CQ
=
2
1
&
CA
CP
=
1
2
QB
CQ
=
1
2
⇒
BC−CQ
CQ
=2
⇒3CQ=2BC
⇒CQ=
3
2
BC
By (1)
AQ
2
=AC
2
+(
3
2
BC)
2
2)9AQ
2
=9AC
2
+4BC
2
−−−−−(A)
In BCP
CA−CP
CP
=2
PB
2
=BC
2
+PC
2
−−CP=
3
2
CA
PB
2
=BC
2
+(
3
2
CA)
2
⇒9PB
2
=9BC
2
+4AC
2
−−−−−(B)
(3) Adding (A)&(B)
9AQ
2
=9BC
2
=13(BC
2
+AC
2
)
⇒9(AQ
2
+BP
2
)=13AB
2
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